Problem 26
Question
Find the amplitude, if it exists, and period of each function. Then graph each function. \(y=6 \sin \frac{2}{3} \theta\)
Step-by-Step Solution
Verified Answer
Amplitude is 6; period is \(3\pi\).
1Step 1: Identifying the Amplitude
The amplitude of a sine function given in the form \( y = a \, \sin \, b\theta \) is the absolute value of \( a \). Here, \( a = 6 \), so the amplitude is \(|6| = 6\).
2Step 2: Determining the Period
The period of a sine function given as \( y = a \, \sin \, b\theta \) is calculated using the formula \( \frac{2\pi}{|b|} \). Here, \( b = \frac{2}{3} \). Thus, the period is \( \frac{2\pi}{\left| \frac{2}{3} \right|} \). This simplifies to \( \frac{2\pi \times 3}{2} = 3\pi \).
3Step 3: Graphing the Function
Plot the sine wave based on the amplitude and period derived. Start at the origin (0,0), rise to the maximum value of 6 at \( \frac{3\pi}{2} \), return to 0 at \( 3\pi \), go to -6 at \( \frac{9\pi}{2} \), and then return back to 0 at \( 6\pi \). This completes one cycle of the sine wave. Repeat this pattern for more cycles and draw smooth curves through these points.
Key Concepts
AmplitudePeriod of a functionGraphing Trigonometric Functions
Amplitude
In trigonometry, the amplitude of a function reflects how far the wave stretches vertically from its central axis. Think of it like the height of a wave crest or the depth of a trough measured from the middle line. For the sine function, which is in the form \[ y = a \, \sin \, b\theta \]Amplitude is identified by the absolute value of \(a\). Thus, if \(a = 6\), then the amplitude is \(|6|\), which is simply 6. The amplitude provides key information about the function: the sine wave will stretch 6 units upwards and 6 units downwards from its original position along the middle line of the graph. This doesn't alter the shape or timing of the sine wave but changes its size.
- A higher amplitude means a taller wave.
- A smaller amplitude indicates a shorter wave.
Period of a function
The period of a trigonometric function defines the length over which the function completes one full cycle before repeating itself. In the context of the sine function, which follows the formula \[ y = a \, \sin \, b\theta \]The period is evaluated using the expression \(\frac{2\pi}{|b|}\). For our function, \(b = \frac{2}{3}\). Plugging this into the formula, we find that the period is \[\frac{2\pi}{\left| \frac{2}{3} \right|} = \frac{2\pi \times 3}{2} = 3\pi\]This means that every \(3\pi\) radians, the wave pattern of the function starts anew. Unlike amplitude, which affects the height of the wave, the period affects its length over the horizontal axis.
- A longer period indicates that the wave takes more space to complete one cycle.
- A shorter period results in tighter cycles.
Graphing Trigonometric Functions
Creating a graph of a trigonometric function requires understanding both the amplitude and period. For example, with the sine function \[ y = 6 \, \sin \frac{2}{3} \theta \]We start the graph at the origin (0,0) because the sine function typically begins there. Then, guided by the amplitude, we note the wave's peak at 6 and its trough at -6, providing visual upper and lower boundaries for our graph. Next, discern the points where the function's value returns to zero, marking them at distinct intervals defined by the period. In this case, the cycle's significant checkpoints include rising to 6 at \(\frac{3\pi}{2}\), returning to 0 at \(3\pi\), dipping to -6 at \(\frac{9\pi}{2}\), and completing the cycle back to 0 at \(6\pi\).
- Ensure symmetry, as sine waves are symmetric around their central axis.
- Smoothly connect these points in an accurate wave pattern.
Other exercises in this chapter
Problem 26
Simplify each expression. \(\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}\)
View solution Problem 26
State the vertical shift, equation of the midline, amplitude, and period for each function. Then graph the function. $$ y=\frac{1}{2} \sin \theta+\frac{1}{2} $$
View solution Problem 27
Solve each equation for all values of \(\theta\). \(2 \cos ^{2} \theta+\cos \theta=0\)
View solution Problem 27
Verify that each of the following is an identity. $$ \frac{1}{\sin x \cos x}-\frac{\cos x}{\sin x}=\tan x $$
View solution